The equation of the tangent at the pointt on the curve x= a(t +sin t),y = a(1-cost) is

Options

A. y=(x-at).tan(t/2)
B. y=(x+at).tan(t/2)
C. y=(x-at).cot(t/2)
D. None of these

Answer

To find the equation of the tangent at a point \( t \) on the curve given by the parametric equations \( x = a(t + \sin t) \) and \( y = a(1 - \cos t) \), we can follow these steps:

### Step 1: Find the coordinates of the point on the curve
The coordinates of the point on the curve at parameter \( t \) are given by:
\[ x_1 = a(t + \sin t) \]
\[ y_1 = a(1 - \cos t) \]

### Step 2: Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
Next, we need to find the derivatives of \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = a(1 + \cos t) \]
\[ \frac{dy}{dt} = a \sin t \]

### Step 3: Find the slope of the tangent line \( m \)
The slope of the tangent line \( m \) can be found using the formula:
\[ m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} \]

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line, we have:
\[ y - y_1 = m(x - x_1) \]
Substituting \( y_1 \), \( m \), \( x_1 \):
\[ y - a(1 - \cos t) = \frac{\sin t}{1 + \cos t} \left( x - a(t + \sin t) \right) \]

### Step 5: Simplify the equation
Now, we can simplify the equation:
\[ y - a(1 - \cos t) = \frac{\sin t}{1 + \cos t} \cdot x - \frac{\sin t}{1 + \cos t} \cdot a(t + \sin t) \]
Multiplying through by \( 1 + \cos t \):
\[ (1 + \cos t)(y - a(1 - \cos t)) = \sin t \cdot x - a \sin t(t + \sin t) \]
Distributing:
\[ y(1 + \cos t) - a(1 - \cos t)(1 + \cos t) = \sin t \cdot x - a \sin t(t + \sin t) \]
This gives us a simplified equation of the tangent line.

### Final Result
The equation of the tangent line at the point \( t \) on the curve is:
\[ y(1 + \cos t) = \sin t \cdot x - a \sin t(t + \sin t) + a(1 - \cos^2 t) \]